Found problems: 1049
An arbitrary triangle ABC is given. There are 2 lines, p and q, that are not parallel to each other and they are not perpendicular to the sides of the triangle. The perpendicular lines through points A, B and C to line p we denote with $ p_a, p_b, p_c $ and the perpendicular lines to line q we denote with $ q_a, q_b, q_c $. Let the intersection points of the lines $ p_a, q_a, p_b, q_b, p_c $ and $ q_c $ with $ q_b, p_b, q_c, p_c, q_a $ and $ p_a $ are $ K, L, P, Q, N $ and $ M $. Prove that $ KL, MN $ and $ PQ $ intersect in one point.
A convex quadrilateral is given. Using a compass and a ruler construct a point such that its projections to the sidelines of this quadrilateral are the vertices of a parallelogram.
(A. Zaslavsky)
$A; B; C; D; E$ and $F$ are points in space such that $AB$ is parallel to $DE$, $BC$ is parallel to $EF$, $CD$ is parallel to $FA$, but $AB \neq DE$. Prove that all six points lie in the same plane.
[i](4 points)[/i]
$ O $ is the center of a parallelogram $ ABCD. $ Let $ G $ on the segment $ OB $ (excluding its endpoints), $ N $ on the line $ DC $ and $ M $ on the segment $ AD $ (excluding its endpoints) such that $ CN>ND, AM=6MD $ and so that there exists a natural number $ n\ge 3 $ such that $ OB=nGO. $ Show that $ G,M,N $ are collinear if and only if
$$ \left( \frac{CN}{ND} -6 \right) (n+1)=2. $$
Let $ABCDE$ be a convex pentagon such that $BC \parallel AE,$ $AB = BC + AE,$ and $\angle ABC = \angle CDE.$ Let $M$ be the midpoint of $CE,$ and let $O$ be the circumcenter of triangle $BCD.$ Given that $\angle DMO = 90^{\circ},$ prove that $2 \angle BDA = \angle CDE.$
[i]Proposed by Nazar Serdyuk, Ukraine[/i]
Of all parallelograms of a given area find the one with the shortest possible longer diagonal.
The diagonals of a parallelogram $ABCD$ meet at point $O$. The tangent to the circumcircle of triangle $BOC$ at $O$ meets ray $CB$ at point $F$. The circumcircle of triangle $FOD$ meets $BC$ for the second time at point $G$. Prove that $AG=AB$.
In $ \triangle{ABC}$, we have $ AB\equal{}1$ and $ AC\equal{}2$. Side $ BC$ and the median from $ A$ to $ BC$ have the same length. What is $ BC$?
$ \textbf{(A)}\ \frac{1\plus{}\sqrt2}{2} \qquad
\textbf{(B)}\ \frac{1\plus{}\sqrt3}{2} \qquad
\textbf{(C)}\ \sqrt2 \qquad
\textbf{(D)}\ \frac{3}{2} \qquad
\textbf{(E)}\ \sqrt3$
The inscribed circle of the triangle $ABC$ touches the sides $BC, CA, AB$ at points $A_1, B_1, C_1$ respectively. The points $P_b, Q_b, R_b$ are the points of the segments $BC_1, C_1A_1, A_1B$, respectively, such that $BP_bQ_bR_b$ is parallelogram. In the same way, the points $P_c, Q_c, R_c$ are the points of the sections $CB_1, B_1A_1, A_1C$, respectively such that $CP_cQ_cR_c$ is a parallelogram. The intersection of the lines $P_bR_b$ and $P_cR_c$ is $T$. Show that $TQ_b = TQ_c$.
Let $ABCD$ be a parallelogram of area $10$ with $AB = 3$ and $BC = 5$. Locate $E$, $F$ and $G$ on segments $\overline{AB}$, $\overline{BC}$ and $\overline{AD}$, respectively, with $AE = BF = AG = 2$. Let the line through $G$ parallel to $\overline{EF}$ intersect $\overline{CD}$ at $H$. The area of the quadrilateral $EFHG$ is
$ \textbf{(A)}\ 4\qquad\textbf{(B)}\ 4.5\qquad\textbf{(C)}\ 5\qquad\textbf{(D)}\ 5.5\qquad\textbf{(E)}\ 6 $
Let $ABCD$ be a trapezoid such that $AB$ is parallel to $CD$, and let $E$ be the midpoint of its side $BC$. Suppose we can inscribe a circle into the quadrilateral $ABED$, and that we can inscribe a circle into the quadrilateral $AECD$. Denote $|AB|=a$, $|BC|=b$, $|CD|=c$, $|DA|=d$. Prove that \[a+c=\frac{b}{3}+d;\] \[\frac{1}{a}+\frac{1}{c}=\frac{3}{b}\]
A quadrilateral $ABCD$ without parallel sides is circumscribed around a circle with centre $O$. Prove that $O$ is a point of intersection of middle lines of quadrilateral $ABCD$ (i.e. barycentre of points $A,\,B,\,C,\,D$) iff $OA\cdot OC=OB\cdot OD$.
Given a triangle $ABC$. A point $X$ is chosen on a side $AC$. Some circle passes through $X$, touches the side $AC$ and intersects the circumcircle of triangle $ABC$ in points $M$ and $N$ such that the segment $MN$ bisects $BX$ and intersects sides $AB$ and $BC$ in points $P$ and $Q$. Prove that the circumcircle of triangle $PBQ$ passes through a fixed point different from $B$.
[b]proposed by Sergej Berlov[/b]
Let $O$ and $P$ be fixed points on a plane, and let $ABCD$ be any parallelogram with center $O$. Let $M$ and $N$ be the midpoints of $AP$ and $BP$ respectively. Lines $MC$ and $ND$ meet at $Q$. Prove that the point $Q$ lies on the lines $OP$, and show that it is independent of the choice of the parallelogram $ABCD$.
Let $ABCD$ be a parallelogram with an acute angle at $A$. Let $G$ be a point on the line $AB$, distinct from $B$, such that $|CG| = |CB|$. Let $H$ be a point on the line $BC$, distinct from $B$, such that $|AB| =|AH|$. Prove that triangle $DGH$ is isosceles.
[asy]
unitsize(1.5 cm);
pair A, B, C, D, G, H;
A = (0,0);
B = (2,0);
D = (0.5,1.5);
C = B + D - A;
G = reflect(A,B)*(C) + C - B;
H = reflect(B,C)*(H) + A - B;
draw(H--A--D--C--G);
draw(interp(A,G,-0.1)--interp(A,G,1.1));
draw(interp(C,H,-0.1)--interp(C,H,1.1));
draw(D--G--H--cycle, dashed);
dot("$A$", A, SW);
dot("$B$", B, SE);
dot("$C$", C, E);
dot("$D$", D, NW);
dot("$G$", G, NE);
dot("$H$", H, SE);
[/asy]
Let $ ABC$ an acute triangle with circumcircle center $ O.$ The line $ (BO)$ intersects the circumcircle again in $ D,$ and the extension of the altitude from $ A$ intersects the circle in $ E.$ Prove that the quadrilateral $ BECD$ and the triangle $ ABC$ have the same area.
Let $ABC$ be an acute triangle and $H$ is orthocenter. Let $D$ be the intersection of $BH$ and $AC$ and $E$ be the intersection of $CH$ and $AB$. The circumcircle of $ADE$ cuts the circumcircle of $ABC$ at $F \neq A$. Prove that the angle bisectors of $\angle BFC$ and $\angle BHC$ concur at a point on $BC.$
On side $BC$ of parallelogram $ABCD$ ($A$ is acute) lies point $T$ so that triangle $ATD$ is an acute triangle. Let $O_1$, $O_2$, and $O_3$ be the circumcenters of triangles $ABT$, $DAT$, and $CDT$ respectively. Prove that the orthocenter of triangle $O_1O_2O_3$ lies on line $AD$.
In parallelogram $ABCD$, the longer side is twice the shorter side. Let $XYZW$ be the quadrilateral formed by the internal bisectors of the angles of $ABCD$. If the area of $XYZW$ is $10$, find the area of $ABCD$
A parallelogram $ABCD$ is given. The excircle of triangle $\triangle{ABC}$ touches the sides $AB$ at $L$ and the extension of $BC$ at $K$. The line $DK$ meets the diagonal $AC$ at point $X$; the line $BX$ meets the median $CC_1$ of trianlge $\triangle{ABC}$ at ${Y}$. Prove that the line $YL$, median $BB_1$ of triangle $\triangle{ABC}$ and its bisector $CC^\prime$ have a common point.
[i](A. Golovanov)[/i]
Let $ABC$ be an equilateral triangle. A point $T$ is chosen on $AC$ and on arcs $AB$ and $BC$ of the circumcircle of $ABC$, $M$ and $N$ are chosen respectively, so that $MT$ is parallel to $BC$ and $NT$ is parallel to $AB$. Segments $AN$ and $MT$ intersect at point $X$, while $CM$ and $NT$ intersect in point $Y$. Prove that the perimeters of the polygons $AXYC$ and $XMBNY$ are the same.
Let $\Omega$ be the circumscribed circle of the acute triangle $ABC$ and $ D $ a point the small arc $BC$ of $\Omega$. Points $E$ and $ F $ are on the sides $ AB$ and $AC$, respectively, such that the quadrilateral $CDEF$ is a parallelogram. Point $G$ is on the small arc $AC$ such that lines $DC$ and $BG$ are parallel. Prove that the angles $GFC$ and $BAC$ are equal.
The vertex $F$ of the parallelogram $ACEF$ lies on the side $BC$ of parallelogram $ABCD$. It is known that $AC = AD$ and $AE = 2CD$. Prove that $\angle CDE = \angle BEF$.
Let $ABCD$ be a convex quadrilateral for which \[ AB+CD=\sqrt{2}\cdot AC\qquad\text{and}\qquad BC+DA=\sqrt{2}\cdot BD.\] Prove that $ABCD$ is a parallelogram.
The vertex $ A$ of the acute triangle $ ABC$ is equidistant from the circumcenter $ O$ and the orthocenter $ H.$ Determine all possible values for the measure of angle $ A.$